Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $n = \dfrac{p - 9}{p - 4} \times \dfrac{p - 7}{p^2 - 16p + 63} $
Solution: First factor the quadratic. $n = \dfrac{p - 9}{p - 4} \times \dfrac{p - 7}{(p - 9)(p - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (p - 9) \times (p - 7) } { (p - 4) \times (p - 9)(p - 7) } $ $n = \dfrac{ (p - 9)(p - 7)}{ (p - 4)(p - 9)(p - 7)} $ Notice that $(p - 7)$ and $(p - 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(p - 9)}(p - 7)}{ (p - 4)\cancel{(p - 9)}(p - 7)} $ We are dividing by $p - 9$ , so $p - 9 \neq 0$ Therefore, $p \neq 9$ $n = \dfrac{ \cancel{(p - 9)}\cancel{(p - 7)}}{ (p - 4)\cancel{(p - 9)}\cancel{(p - 7)}} $ We are dividing by $p - 7$ , so $p - 7 \neq 0$ Therefore, $p \neq 7$ $n = \dfrac{1}{p - 4} ; \space p \neq 9 ; \space p \neq 7 $